Extract Kth Character From The Rope Tree

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TreesRecursionStrings

A rope tree is a binary tree that represents a string. Each node in the tree has a weight, which is the length of the string it represents. Each leaf node represents a single character. The internal nodes represent the concatenation of the strings represented by their children.

Given the root of the rope tree and an integer k, return the k^th character of the string represented by the tree.

Note:

Assume that k is 1-indexed.
It is guaranteed that k is valid, which means k is in the range [1, The length of the string represented by the tree].

Example 1:

Input: root = [5,3,2,"abcde","st","fg"], k = 3
Output: "c"
Explanation: The string represented by the tree is "abcde" + "st" + "fg" = "abcdeSTfg", so the 3^rd character is "c".

Example 2:

Input: root = [4,1,3,"i","g","ide"], k = 1
Output: "i"
Explanation: The string represented by the tree is "i" + "g" + "ide" = "igide", so the 1^st character is "i".

Example 3:

Input: root = [6,4,2,[1,1,2,"ef","gh","ij"],"kl"], k = 5
Output: "h"
Explanation: The string represented by the tree is ("ef" + "gh" + "ij") + "kl" = "efghijkl", so the 5^th character is "h".

Constraints:

The number of nodes in the tree is in the range [1, 1000].
1 <= Node.weight <= 1000
If the node is a leaf node, the length of the string it represents is equal to the node's weight.
Each node has either 0 or 2 children.
If the node is not a leaf node, the node's weight is equal to the sum of the weights of its children.
1 <= k <= The length of the string represented by the tree
The string consists of only lowercase and uppercase English letters.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the structure of the Rope Tree? Specifically, how are the nodes defined (e.g., do they store lengths, substrings, or both)?
What range of values can 'k' have? Is it 0-indexed or 1-indexed?
If 'k' is out of bounds (i.e., larger than the length of the string represented by the Rope Tree), what should be returned (e.g., null, an error string, or should an exception be thrown)?
Can the Rope Tree be empty, and if so, what should be returned?
Are the leaf nodes guaranteed to contain only single characters or can they hold substrings?

Brute Force Solution

Approach

Think of the rope tree as a really long string that's been broken down into smaller pieces. To find the Kth character, the brute force way is to simply walk through all the pieces, adding up their lengths until you reach the piece containing the character you want. Then you just pick out that character.

Here's how the algorithm would work step-by-step:

Start at the beginning of the rope tree.
Go through each piece of the tree one by one.
Keep a running total of the lengths of the pieces you've gone through so far.
If the running total is less than K, move to the next piece.
If the running total is greater than or equal to K, you've found the piece that contains the Kth character.
Figure out how far into that piece the Kth character is.
That's your character!

Code Implementation

def extract_kth_character_brute_force(rope_tree, k_index):
    running_total_length = 0

    for text_segment in rope_tree:
        # Check if the current segment contains the kth character

        if running_total_length + len(text_segment) >= k_index:

            # Calculate the index of the character within the segment
            index_within_segment = k_index - running_total_length

            # Return the character at the calculated index
            return text_segment[index_within_segment]

        # Update the running total length for the next iteration
        running_total_length += len(text_segment)

    # If k_index is out of bounds, return None
    return None

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the pieces of the rope tree until it finds the piece containing the Kth character. In the worst case, it might have to traverse all n pieces of the rope tree. Each piece only requires a constant time operation to calculate the running total and compare it with K. Therefore, the time complexity is directly proportional to the number of pieces, n, resulting in O(n).

Space Complexity

O(1) – The provided algorithm iterates through the pieces of the rope tree, maintaining a running total of lengths. The only auxiliary space required is for storing a few scalar variables, such as the running total and the current piece index. The number of such variables is constant and does not depend on the size of the rope tree (number of pieces, N). Therefore, the space complexity is O(1).

Optimal Solution

Approach

Imagine the rope tree is like a very clever way to store a long string of text. To find a specific character, we don't need to look at every single character in the whole thing. We can use the structure of the tree to jump directly to the part that contains the character we want.

Here's how the algorithm would work step-by-step:

Start at the very top of the rope tree.
Look at the size (number of characters) of the left branch of the tree.
If the character we're looking for is within that size, then it's in the left branch. If not, it's in the right branch.
If it's in the right branch, reduce our target character number by the size of the left branch.
Repeat the process on the branch where our character lies.
Continue doing this until we reach a small piece of text (a leaf of the tree).
The character we're looking for will be in that small piece of text, at the position we've narrowed down.

Code Implementation

class RopeNode:
    def __init__(self, data='', left=None, right=None, weight=0):
        self.data = data
        self.left = left
        self.right = right
        self.weight = weight

def extract_kth_character(root, k_index):
    current_node = root
    while current_node:
        # If it's a leaf node, the character is within its data.
        if not current_node.left and not current_node.right:
            return current_node.data[k_index]

        # If there is a left child, compare k with its weight
        if current_node.left:
            if k_index < current_node.weight:

                current_node = current_node.left
            else:
                # Kth char is in the right subtree
                k_index -= current_node.weight
                current_node = current_node.right
        else:
            # If there is no left child, traverse to the right.
            current_node = current_node.right
    return None

Big(O) Analysis

Time Complexity

O(H) – The algorithm traverses the rope tree from the root to a leaf. At each node, it compares the target index k with the size of the left subtree to determine which subtree to explore next. The height of the tree (H) dictates the number of steps required to reach a leaf. Therefore, the time complexity is directly proportional to the height of the tree.

Space Complexity

O(log N) – The primary auxiliary space usage comes from the recursion stack. The maximum depth of the recursion is determined by the height of the rope tree. In a balanced rope tree, the height is logarithmic with respect to the total number of characters N stored in the tree, where N represents the total number of characters in the conceptual string. Therefore, the space used by the recursion stack is O(log N).

Edge Cases

Case	How to Handle
Null rope tree root	Return null or throw an exception as appropriate for an empty tree.
K is less than 1	Return an error, throw an exception, or define the behavior (e.g., return first character) and document it.
K is greater than the total length of the rope tree	Return an error, throw an exception, or define the behavior (e.g., return null character) and document it.
A node with empty string and non-zero weight	Treat as an error as the weight should represent the string length and handle appropriately.
Integer overflow when calculating or comparing sums (weight)	Use a larger data type (e.g., long) or check for potential overflows before calculations.
Rope tree is highly unbalanced (e.g., long chain of single characters)	The algorithm should handle this case, but performance may degrade; consider balancing tree structure if modifications are allowed.
Node contains extremely long string	Accessing the string at an offset may cause an exception (string index out of bound) or slow down access and should be considered when choosing data structures.
Negative weight encountered in the Rope tree	Throw an exception, or handle as an invalid state, since weight should be non-negative.

Null rope tree root

How to Handle:

Return null or throw an exception as appropriate for an empty tree.

K is less than 1

How to Handle:

Return an error, throw an exception, or define the behavior (e.g., return first character) and document it.

K is greater than the total length of the rope tree

How to Handle:

Return an error, throw an exception, or define the behavior (e.g., return null character) and document it.

A node with empty string and non-zero weight

How to Handle:

Treat as an error as the weight should represent the string length and handle appropriately.

Integer overflow when calculating or comparing sums (weight)

How to Handle:

Use a larger data type (e.g., long) or check for potential overflows before calculations.

Rope tree is highly unbalanced (e.g., long chain of single characters)

How to Handle:

The algorithm should handle this case, but performance may degrade; consider balancing tree structure if modifications are allowed.

Node contains extremely long string

How to Handle:

Accessing the string at an offset may cause an exception (string index out of bound) or slow down access and should be considered when choosing data structures.

Negative weight encountered in the Rope tree

How to Handle:

Throw an exception, or handle as an invalid state, since weight should be non-negative.